Let's say an object is thrown from Earth with a speed of v at an angle $\theta$. An ellipse is formed with the center of Earth as its foci(given in the figure).
We know that $v_x^2=2GM(\frac{1}{r_x}-\frac{1}{2a})$ so:$$
v_a^2=2GM(\frac{1}{r_a}-\frac{1}{2a}),v_p^2=2GM(\frac{1}{r_p}-\frac{1}{2a}),v^2=2GM(\frac{1}{R}-\frac{1}{2a})$$(where $r_a$ is the apogee,$r_p$ is the perigee, R is the radius of Earth and a is the semi-major axis 2a=$r_a+r_p$)also the max height of projectile $\frac{v^2sin^2\theta}{2g}=r_a-R$
. To solve for the equation of the ellipse we need to first solve for $r_a,r_p$
($v_a=vcos\theta$). By conservation of angular momentum($mv_xr_xsinx=constant$):
$$v_a*r_a=v_p*r_p=R*vsin\theta$$
$$2GM(\frac{r_a*r_p}{r_a+r_p})=R^2v^2sin^2\theta$$
The problem arises here :
- Substituting $v^2sin^2\theta$ from projectile motion equation:$$2GM(\frac{r_a*r_p}{r_a+r_p})=R^2*2g(r_a-R)=R^2*\frac{2GM}{R^2}(r_a-R)=2GM(r_a-R)$$$$\frac{1}{r_p}=\frac{1}{r_a-R}-\frac{1}{r_a}$$ 2.$v_a^2=v^2cos^2\theta=2GM(\frac{1}{r_a}-\frac{1}{2a}),v^2=2GM(\frac{1}{R}-\frac{1}{2a})$ therefore $v^2sin^2\theta=2GM(\frac{1}{R}-\frac{1}{r_a})$: $$2GM(\frac{r_a*r_p}{r_a+r_p})=R^2(2GM(\frac{1}{R}-\frac{1}{r_a}))$$$$\frac{r_a*r_p}{r_a+r_p}=\frac{1}{R}-\frac{1}{r_a}$$ which is not the same as the previous equation(My guess is that the equation of projectile motion cannot be applied and the 2nd process is correct instead of the 1st one).[Now obviously I could have done $\frac{v^2cos^2\theta}{2GM} =\frac{r_p}{r_a(r_a+r_p)}$ and $\frac{1}{r_a}=\frac{1}{R}-\frac{v^2sin^2\theta}{2GM}$ and then we can solve for $r_p$ but I tried solving it by the process above.]
